Lagrangian Multipliers to find maximum and minimum values

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SUMMARY

The discussion centers on the application of Lagrangian multipliers to find maximum and minimum values of functions constrained by other functions. The key concept is that the gradients of the functions involved, denoted as ∇f and ∇g, must be scalar multiples of each other at the extrema. However, the user raises a critical point about scenarios where the gradients are not parallel, suggesting that the constraint may not effectively limit the function, leading to confusion about the location of extrema. The example provided illustrates the mathematical process using the function f(x,y) = x² + y² with the constraint g(x,y) = y - x.

PREREQUISITES
  • Understanding of multivariable calculus, specifically gradient vectors.
  • Familiarity with the concept of level curves in three-dimensional space.
  • Knowledge of Lagrange multipliers and their application in optimization problems.
  • Ability to manipulate equations involving gradients and constraints.
NEXT STEPS
  • Study the geometric interpretation of Lagrange multipliers in optimization.
  • Explore examples of Lagrangian multipliers with non-linear constraints.
  • Learn about the implications of non-parallel gradients in optimization problems.
  • Investigate alternative optimization methods for constrained functions.
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Students and professionals in mathematics, engineering, and economics who are learning optimization techniques, particularly those interested in applying Lagrangian multipliers to real-world problems.

cooev769
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I'm just learning this theory and the maths is really trivial but the theory is slightly confusing me.

I understand that if we have some function z=f(x,y) and we graph this on a three dimensional set of axis we will have some surface, we can then extend this by creating level curves in the x, y plane for differing values of z.

We then set up a function g(x,y)=k such that k is a constant, we call this our constraint curve and it looks something as shown in the link I have provided.

http://en.wikipedia.org/wiki/File:LagrangeMultipliers2D.svg

I understand how the theory that the maximum/minimum value will occur when the grad of function f and the grad of function g are scalar multiples of each other. But imagine for a second that the constraint curve was a straight line and it went straight through the centre of the circle created by the level curve of the function f. In that case the maximum value would be at the centre of this level curve, but to my knowledge at no time were the gradients parallel.

Or imagine a sloping flat surface sloping up. If the constraint curve started off running up the slope then for a second ran tangent to one of the level curves and then ran at some angle up this surface and stopped, then the point at which the gradients were scalar multiples only in the middle of the curves, so the supposed maximum/minimum occurred at a random place. If you could clarify this it would be extremely helpful thanks!
 
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cooev769 said:
I'm just learning this theory and the maths is really trivial but the theory is slightly confusing me.

I understand that if we have some function z=f(x,y) and we graph this on a three dimensional set of axis we will have some surface, we can then extend this by creating level curves in the x, y plane for differing values of z.

We then set up a function g(x,y)=k such that k is a constant, we call this our constraint curve and it looks something as shown in the link I have provided.

http://en.wikipedia.org/wiki/File:LagrangeMultipliers2D.svg

I understand how the theory that the maximum/minimum value will occur when the grad of function f and the grad of function g are scalar multiples of each other. But imagine for a second that the constraint curve was a straight line and it went straight through the centre of the circle created by the level curve of the function f. In that case the maximum value would be at the centre of this level curve, but to my knowledge at no time were the gradients parallel.
So like: "Minimize f(x,y)= x^2+ y^2 subject to the constraint y= x"? The constraint can be written g(x,y)= y- x so \nabla f= 2x\vec{i}+ 2y\vec{j} and \nabla g= -vec{i}+ vec{j}. The Lagrange multiplier method says that the max or min will be where 2x= -\lambda and 2y= \lambda.

Eliminate \lambda by dividing one equation by the other: x/y= -1 or y= -x. That, together with y= x, the constraint, give y= x= 0. The two gradients are never parallel but if one is 0 that will also satisfy \nabla g= \lambda\nabla f.

Another way of putting this is that if the two gradients are never parallel, then the "constraint" doesn't really constrain anything. The max or min will be where the gradient of the given function is 0.

Or imagine a sloping flat surface sloping up. If the constraint curve started off running up the slope then for a second ran tangent to one of the level curves and then ran at some angle up this surface and stopped, then the point at which the gradients were scalar multiples only in the middle of the curves, so the supposed maximum/minimum occurred at a random place. If you could clarify this it would be extremely helpful thanks!
 

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